Thread: What are the Zero's of this function?

the problem is {x}^{4} - 16

I get;
{x}^{4} - {2}^{4} = {(x+2)}^{2} + {(x-2)}^{2}

for zeros of 2 and -2 each of multiplicity of 2.

However the book states zeros of 2 and -2 each of multiplicity 1. Could someone be so kind as to explain to me what I have done wrong?

Originally Posted by bkbowser

the problem is {x}^{4} - 16

I get;
{x}^{4} - {2}^{4} = {(x+2)}^{2} + {(x-2)}^{2}

for zeros of 2 and -2 each of multiplicity of 2.

However the book states zeros of 2 and -2 each of multiplicity 1. Could someone be so kind as to explain to me what I have done wrong?

You can applied the difference of square incorrectly.



Now use the difference of squares again!

Originally Posted by TheEmptySet

You can applied the difference of square incorrectly.



Now use the difference of squares again!

Is this it? It does give zeros of 2 and -2 each of multiplicity 1. And since the question just asked about real zeros the other 2 wouldn't count?

Originally Posted by bkbowser

Is this it? It does give zeros of 2 and -2 each of multiplicity 1. And since the question just asked about real zeros the other 2 wouldn't count?

yes

Another way to solve is to write it as . Taking the square root of each side we have and . From , taking the square root again, x= 2 and x= -2. There are no real number x such that .